In any $$ riangle A B C, \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} $$ equals to

Question

In any $$ 	riangle A B C, \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} $$ equals to

Aicharya · Accepted Answer

$\begin{aligned} & \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} \ & =\frac{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}{4 b^2 c^2} \ & =\frac{16[s(s-a)(s-b)(s-c)]}{4 b^2 c^2} \quad(\because a+b+c=2 s) \ & =\frac{4 \Delta^2}{b^2 c^2}=\left(\frac{2 \Delta}{b c}ight)^2=\sin ^2 A\left[\because \sin A=\frac{2 \Delta}{b c}ight]\end{aligned}$

In any $$ \triangle A B C, \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} $$ equals to

Solution